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A circular disc D₁ of mass M and radius R has two identical discs D₂ and D₃ of the same mass M and radius R attached rigidly at its opposite ends (see figure). The moment of inertia of the system about the axis OO', passing through the centre D₁, as shown in the figure, will be :

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$3 {MR}^{2}$

${MR}^{2}$

$\frac{2}{3} {MR}^{2}$

$\frac{4}{5} {MR}^{2}$

answer is A.

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Detailed Solution

From the figure,

$I = I_{1} + I_{2} + I_{3} = I_{1} + 2 I_{2}$

$I = \frac{{MR}^{2}}{2} + 2 (\frac{{MR}^{2}}{4} + {MR}^{2})$

$I = \frac{{MR}^{2}}{2} + \frac{{MR}^{2}}{2} + 2 {MR}^{2} = 3 {MR}^{2}$

Hence the correct answer is $3 {MR}^{2} .$

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